Calculate the cell potential for Ag_{(s)}|Ag^ | vedclass

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Question

(A) The cell reaction is: $Ag_{(s)} + Ag^{+}_{(0.1 \ M)} \rightarrow Ag^{+}_{(0.01 \ M)} + Ag_{(s)}$.Using the Nernst equation at $298 \ K$:$E_{cell}

Vedclass · Accepted Answer

$0.0592$

Vedclass · Answer

(A) The cell reaction is: $Ag_{(s)} + Ag^{+}_{(0.1 \ M)} ightarrow Ag^{+}_{(0.01 \ M)} + Ag_{(s)}$.Using the Nernst equation at $298 \ K$:$E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \frac{[Ag^{+}]_{anode}}{[Ag^{+}]_{cathode}}$.Here,$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.80 \ V - 0.80 \ V = 0 \ V$.$n = 1$ (number of electrons transferred).$E_{cell} = 0 - \frac{0.0591}{1} \log \frac{0.01}{0.1}$.$E_{cell} = -0.0591 	imes \log(0.1) = -0.0591 	imes (-1) = 0.0591 \ V \approx 0.0592 \ V$.

Calculate the cell potential for $Ag_{(s)}|Ag^{+} \, (0.01 \ M)||Ag^{+} \, (0.1 \ M)|Ag_{(s)}$ at $298 \ K$. (in $V$)

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